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learning, the parity problem, and the statistical query model. Journal of the ACM 50: 506-519, 2003. [32] S. C. Brubaker and S. Vempala. Isotropic PCA and affine-invariant clustering. In FOCS, pages 551–560, 2008. [33] E. Candes and B. Recht. Exact matrix completion via convex optimization. Foundations of Computati...
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1):33–61, 1998. [40] A. Cohen, W. Dahmen and R. DeVore. Compressed sensing and best k-term approximation. Journal of the AMS, pages 211–231, 2009. [41] J. Cohen and U. Rothblum. Nonnegative ranks, decompositions and factor­ izations of nonnegative matrices. Linear Algebra and its Applications, pages 149–168, 1993. ...
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2973, 2007. [49] S. Deerwester, S. Dumais, T. Landauer, G. Furnas and R. Harshman. Indexing by latent semantic analysis. JASIS, pages 391–407, 1990. [50] A.P. Dempster, N.M. Laird, and D.B. Rubin. Maximum likelihood from in­ complete data via the EM algorithm. Journal of the Royal Statistical Society Series B, page...
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all trees. I. Random Structures and Algorithms 14:153-184, 1997. [58] M. Fazel. Matrix Rank MInimization with Applications. PhD thesis, Stanford University, 2002. [59] U. Feige and R. Krauthgamer. Finding and certifying a large hidden clique in a semirandom graph. Random Structures and Algorithms, pages 195–208, 20...
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few coefficients in any basis. arxiv:0910.1879, 2009. [68] D. Gross, Y-K Liu, S. Flammia, S. Becker and J. Eisert. Quantum state to­ mography via compressed sensing. Physical Review Letters, 105(15), 2010. [69] V. Guruswami, J. Lee, and A. Razborov. Almost euclidean subspaces of fn 1 via expander codes. Combinatoric...
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. Hummel and B. C. Gidas. Zero crossings and the heat equation. Courant Institute of Mathematical Sciences TR-111, 1984. [78] R. Impagliazzo and R. Paturi. On the complexity of k-SAT. J. Computer and System Sciences 62(2):pp. 367–375, 2001. [79] A. T. Kalai, A. Moitra, and G. Valiant. Efficiently learning mixtures of...
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and H. Seung. Algorithms for non-negative matrix factorization. In NIPS, pages 556–562, 2000. [87] S. Leurgans, R. Ross and R. Abel. A decomposition for three-way arrays. SIAM Journal on Matrix Analysis and Applications, 14(4):1064–1083, 1993. [88] M. Lewicki and T. Sejnowski. Learning overcomplete representations....
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–375, 2005. [97] B. Olshausen and B. Field. Sparse coding with an overcomplete basis set: A strategy employed by V1? Vision Research, 37(23):331–3325, 1997. [98] C. Papadimitriou, P. Raghavan, H. Tamaki and S. Vempala. Latent semantic indexing: A probabilistic analysis. JCSS, pages 217–235, 2000. [99] Y. Pati, R. ...
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105] R. A. Redner and H. F. Walker. Mixture densities, maximum likelihood and the EM algorithm. SIAM Review, 26(2):195-239, 1984. [106] J. Renegar. On the computational complexity and geometry of the first-order theory of the reals. Journal of Symbolic Computation, pages 255-352, 1991. [107] A. Seidenberg. A new dec...
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Y 123 [115] S. Vavasis. On the complexity of nonnegative matrix factorization. SIAM Journal on Optimization, pages 1364-1377, 2009. [116] S. Vempala, Y. Xiao. Structure from local optima: Learning subspace juntas via higher order PCA. Arxiv:abs/1108.3329, 2011. [117] S. Vempala and G. Wang. A spectral algorithm f...
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18.725 Algebraic Geometry I Lecture 4 Lecture 4: Grassmannians, Finite and Affine Morphisms Remarks on last time 1. Last time, we proved the Noether normalization lemma: If A is a finitely generated k-algebra, then, A contains B ∼= k[x1, . . . , xn] (free subring) such that A is a finitely generated B-module. Question: Whe...
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a = , b = , (a = b2) X X Y ∼= A1 = U1 Note that U1 ∩ U2 ∼= A1 \ {0}. By changing coordinates, we can take the degree 2 curve in P2 to be X 2 + Y 2 = Z 2. Connect points in a quadric to a fixed point. In practice, we can work with the point (1 : 0 : 1). We identify the set of all lines through a given point with P1. Taki...
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X = n (cid:91) i=1 Xi, where the Xi are closed irreducible subsets of X. Without loss of generality, we can assume that none of the Xi are a subset of another. Then, Xi is not a subset of (cid:91) Xj (follows from j =i j ∩ Xi. Since every irreducibility). Otherwise, we would have that Xi is a union of proper closed sub...
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ible. Thus, we must either have f = 0 or g = 0 and A has no zerodivisors. Conversely, suppose that Spec A is not irreducible. Let X = Spec A. Then, we can write X = Z1 ∪ Z2, where Z1, Z2 (cid:40) X are proper closed subsets of X. Since proper closed subsets correspond to nonzero ideals, we can pick nonzero f ∈ IZ1 and ...
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��nes a projective algebraic variety. The embedding of Gr(k, n) into projective space is defined by W (cid:55)→ the line W ⊂ V . k (cid:94) k (cid:94) Claim: This map realizes Gr(k, n) as a closed subvariety in P (cid:33) (cid:32) k (cid:94) V (n) = P k −1. Example 2. Consider the case n = 4 and k = 2. These are lines i...
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or p. 42 – 44 (in 3rd edition) in Section 1.4.1 (“Closed Subsets of Projective Space”) of Basic Algebraic Geometry 1 by Igor Shafarevich. Finite and affine morphisms Definition (cid:91) Y = Ui where the Ui are affine open pieces such that the f −1(Ui) ⊂ X are affine. 1. A morphism of algebraic varieties f : X −→ Y is called a...
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proof of the lemma (use similar ideas as last time) (compare with Lemma 2.4.3 on p. 19 of Kempf). Proof. Let f : X −→ Y be a finite map. We can assume X and Y are affine (statement local on line). Since the composition of two finite maps is finite, we can also assume that Z = X. Write X = Spec A and Y = Spec B and let I = A...
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tf ) −→ Spec A. Example 4. The morphism A2 \ {0} −→ A2 is not affine. This is similar to an exercise in the homework (Problem 3 of Problem Set 1). It actually follows from this and the exactness of localization. Let U ⊂ A2 be an open neighborhood of 0 such that U = A2 \ Zf for some f . Since k[U ] = k[U \ {0}], U \ {0} i...
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is only for a given chain. Here are some facts about the dimension of a Noetherian topological space: • dim An = n • If X = n (cid:91) i=1 Ui, then dim X = max dim Ui. i • If f : X −→ Y is a finite and surjective morphism, then dim X = dim Y . 4 (cid:54) (cid:54) (cid:54) MIT OpenCourseWare http://ocw.mit.edu 18.725 Al...
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Lecture 4 Removable Singularity Theorem Theorem 1 Let u be harmonic in Ω \ {x0}, if � u(x) = | 2−n) o( x − x0 | | | o(ln x − x0 ) , n > 2, , n = 2 as x → x0, then u extends to a harmonic function in Ω. Proof: Without loss of generality, we can assume Ω = B(0, 2), then u|∂B(0,1) is contin­ uous. Thus by Poisson ...
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(x) ≤ u(x), ∀x ∈ B1(0) \ {0}, thus v(x) = u(x), ∀x ∈ B1(0) \ {0}. Now we can define u(0) = v(0), and extend u to be a harmonic function on B(0, 1), thus a harmonic function on Ω = B(0, 2). � Example This gives an example of Dirichlet problem that is NOT solvable: Take Ω = B(0, 1) \ {0}, then ∂Ω = ∂B(0, 1) ∪ {0}. C...
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degree k restricted to Sn−1, then ΔSn−1 B(θ) = −k(k + n − 2)B(θ). Remark 1 Let Pk be the set of homogeneous polynomials of degree k on Rn , Hk be the set of harmonic homogeneous polynomials of degree k on Rn, then It’s not hard to prove P k = H 2 k ⊕ r P k−2. dimPk = (k + n − 1)! , k!(n − 1)! so dimHk = (k + ...
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− n + 2, if k = 0, then p = 2 − n and B(θ) = constant, thus u = c · r2−n, which is the fundamental solution. if k > 0, then p < 2 − n, note that B(θ) is defined on the compact set Sn−1, thus B is bounded, so u grows faster than the fundamental solution near the origin. From above we get a degree gap of harmonic func...
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Engineering Systems Doctoral Seminar ESD.83 – Fall 2011 Class 1 Faculty: Chris Magee and Joe Sussman TA: Rebecca Kaarina Saari Guest: Professor Joel Moses, Institute Professor, (EECS and ESD) © 2010 Chris Magee and Joseph Sussman, Engineering Systems Division, Massachusetts Institute of Technology 1 Session 1: Overvie...
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acy: Understanding of core concepts and principles – base level of literacy on the various aspects of engineering systems  Inter-disciplinary capability: The capability to reach out to adjacent fields in a respectful and knowledgeable way and the ability to engage with other ES scholars in assessing the importance ...
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-The Incorporation of Social Science and Engineering into Research and Problem Solving  Guest: Joel Moses Class 2: How do we know what we know? - The Generation of Cumulative Knowledge in a Field  Guest: David Kaiser Class 3: Modeling Paradigms: Useful models and various modeling approaches  Guest: Oli de Weck C...
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Technology 9 Assignment Summary (syllabus)  1. Observations, Data Sources and Data Reduction Assignment (no more than 1000 word paper, 10% of the total) Students will be expected to select and read an NBER working paper from a faculty-provided list and to prepare a no more than 1000 word paper, performing a critic...
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10% of the total) [assigned Session 3, due due on Session 10] © 2009 Chris Magee, Engineering Systems Division, Massachusetts Institute of Technology 12 Assignment Summary (syllabus) 4  8. Developing a Well-Posed Research Question*** (750- word paper, 10% of the total)[Session 8 - due Session 12]  9. In-Depth Paper ...
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to bridge- specialization effects & ES  Are there significant differences among different sciences? © 2009Chris Magee and Joseph Sussman, Engineering Systems Division, Massachusetts Institute of Technology 15 Relationships among fields of knowledge 2  There are significant differences in different fields with “h...
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-based theory across disciplines..”  Wilson argues that each discipline should be “consilient” with established science in other disciplines.  He distinguishes between examples comparing consilience by reduction (dissect a phenomenon into its components) and consilience by synthesis (predicting higher-order phe...
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Notes -on double integrals. (Read 11.1-11.5 of Apostol.) Just as for the case of a single integral, we have the following condition for the existence of a double integral: Theorem 1 (Riemann condition). Suppose f -is defined -on Q = [arb] x [c,d]. Then f -is integrable -on Q - -if and only -if given any E > 0, th...
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Q, - - and i f f - and g - a r e i n t e g r a b l e TO prove t h i s theorem, one f i r s t v e r i f i e s t h e s e r e s u l t s f o r s t e p f u n c t i o n s (see 1 1 . 3 ) , and t h e n u s e s t h e Riemann condi- t i o n t o prove them f o r g e n e r a l i n t e g r a b l e f u n c t i o n s . The p r o...
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to this partition. Furthermore, one adds the earlier inequalities to obtain Finally, we compute this computation uses the fact that linearity has already been proved for step functions. ~ h u s JJQ (f + g) exists. TO calculate this integral, we note that by definition. ' Then here again we use the linearity of th...
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the double integral ,b f * Choose step functions s (x,y) and t(x,y) , defined on Q, such that s(x,y) C f (x,y) t(x,y), and his w e can do because f exists. For convenience, choose s and t so they are constant on the ?artition lines. (This does not affect their double integrals.) Then the one-dimen- sional integr...
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, for all y in [c,d] , .- Thus S and T are step functions lying beneath and above A, respectively, Furthermore (see p. 3561, so that .. j ) It f o l l o w s t h a t A ( y ) d y exists, by t h e Riemann c o n d i t i o n . Now t h a t w e know A(y) i s i n t e g r a b l e , w e can conclude from a n e a r l ...
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b a s i c q u e s t i o n s , t h e one concerning t h e e x i s t e n c e of t h e double i n t e g r a l . W e r e a d i l y prove t h e f o l l o w i n g : i Theorem 4 . - The i n t e g r a l on t h e r e c t a n g l e Q. continuous - - f e x i s t s - i f f - i s ~ r o o f . A l l one needs i s t h e smal...
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. ( 3 ) A v e r t i c a l l i n e segment h a s c o n t e n t zero. ( 4 ) A s u b s e t o f a set o f c o n t e n t z e r o h a s c o n t e n t zero. (5) A f i n i t e union o f sets o f c o n t e n t z e r o h a s c o n t e n t zero. (6) The graph o f a c o n t i n u o u s f u n c t i o n y = $(x); a < x <.b i ...
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less than E ' . Consider t h e r e c t a n g l e s f o r i = l,...,n. ( 1 - Q x i 1 < E whenever x 1 The t o t a l a r e a of t h e r e c t a n g l e s Ai They cover t h e graph of Q, because i s i n the i n t e r v a l ,xi] . ( x ~ - x ) 2 € ' = 2 c 1 ( b i-1 i=l e q u a l s - a ) . T h i s number e q ...
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a i n Note t h a t t h i s lemma does n o t s t a t e merely t h a t D i s c o n t a i n e d -i n t h e union o f f i n i t e l y many s u b r e c t a n g l e s of t h e par- t i t i o n having t o t a l a r e a l e s s t h a n E, b u t t h a t t h e sum of t h e a r e a s o f -a l l t h e s u b r e c t a n g l e ...
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t i t i o n . Note t h a t by c o n s t r u c t i o n , t h e r e c t a n g l e Aj; i s p a r t i , t i o n e d by P , s o t h a t it i s a union of s u b r e c t a n g l e s Q i j of P. Now i f a s u b r e c t a n g l e Qij t h e n I n t A1; f o r some k t s o that it a c t u a l l y c o n t a i n s a p o i n...
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- If f - is bounded - on Q, and is continuous -on Q except on - - - -a set of content - -zero, then I/', f exists. Proof. S t e ~ 1.We prove a preliminary result: Suppose that given e > 0, there exist functions g and h that are integrable over Q, such that and g(x) I f(x) l h(x) for x in Q Then f is integrable ...
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of D, set g(x) = f(x) = h(x) for x E Q. .. Do this for each such subrectangle. Then for any other x in Q, set 1J g(x) = -M and h(x) = M. T h e n g S f S h o n Q . Now g is integrable over each subrectangle Q.. that does not contain a point of D, 1J since it equals the continuous function f there. And g is inte...
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of D, set g(x) = f(x) = 0 and h(x) = f(x) = 0 on Q. .. Do this for each 1J such subrectangle. For any other x in Q, set T h e n g S f s h o n Q . g(x) = -M and h(x) = M. I Now g and h are step functions on Q, because they are constant on the interior of each subrectangle Q. .. We compute 1J JJQ h = M (1(area...
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t h e s p e c i a l f d e f i n e d on a bounded f o r a f u n c t i o n c a s e where S i s a r e g i o n of Types I o r 11. We d i s c u s s h e r e t h e g e n e r a l c a s e . F i r s t , w e prove t h e f o l l o w i n g b a s i c e x i s t e n c e theorem: Theorem 9. - L e t S be - - a bounded s e t i n t...
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s o continuous a t each p o i n t xl of t h e e x t e r i o r of S, because it equals z e r o on an open b a l l about xl. The only p o i n t s where can' fail t o b e continuous are p o i n t s of ths boundary of S , and this set, by assumption, has c o n t e n t zero. Hence f! e x i s t s . El Q -Note: Adjoi...
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both integrals e x i s t . i (c) Additivity. Let S = SL U S2. If S1 n S2 has content zero, then provided the right side exists. Proof. (a) Given f, g defined on S, let 2, g equal f t gr respectively, on S and equal 0 otherwise. Then cl + dg equals cf + dg on S and 0 otherwise. Let Q be a rectangle containing S...
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regions Sl and S2 that i intersect in a set of content zero. Since S1 is of type I and S2 is of type 11, we can compute the integrals I[ and {I s, f by iterated integration. We add the results to s,A. f obtain -Area. We can now construct a rigorous theory of area. We already have defined the area of the rect...
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Bd S) . Proof. Let Q be a rectangle containing S and T. Let is(x) = 1 for x E S = 0 for x ft S. Define FT similarly. (1) If S is contained in T, then $ (x) C lT(x) . Then by the comparison theorem, area s = Ifs 1 = 11, L < /I,% = j'b I. = area T. (2) Since 0 < 1, we have by the comparison theorem, 0 = 11, 0....
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SUT S T (4) Since the part of S not in Int S lies in Bd S, it has content zero. Then additivity implies that area S = area(1nt S) + area (S - Int S) = area (Int S) . A similar remark shows that area ( S u Bd SJ = area (Int S ) + area(Bd S) = area (Int S ) . 1 Remark. L e t S be a bounded s e t i n t h e p...
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t h e numbers a ( P ) , a s P and d e f i n e t h e o u t e r -a r e a o f S t o b e t h e infemum of t h e numbers A ( P ) . If t h e i n n e r a r e a and o u t e r a r e a of S a r e e q u a l , t h e i r common v a l u e i s c a l l e d t h e -a r e a of S. W e leave it a s a ( n o t t o o d i f f i c u l t ) e...
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r a l . I Eh ercises 1. Show t h a t i f ISS 1 e x i s t s , then Bd S hz-s content zero. [Hint: Chwse Q s o t h a t S C Q . Since S& IS e x i s t s , t h e r e a r e functions s and t t h a t a r e s t e p functions r e l a t i v e t o a p a r t i t i o n P of Q, such t h a t s <, Is jt o Q and [$ ( t - s ) <...
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r f a c e s z = xy and z = 0 and Let Q denote the rectangle [0,1] x [0,1] in the following exercises. @(a) Let f(x,y) = l/(y-x) if x # y, f(x,y) = 0 Does JJQf exist? if x = y. (b) Let g(x,y) = sin (l/(y-x)) if x # y, Does JJQg exist? @ ~ e t f(x,y) = 1if x = 112 and y is rational, ,- f(x,y) = 0 otherwise S...
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T Y R E B I L 4 199 Source: Wikipedia Jacob (James) Bernoulli (1654–1705) For Bernoulli Trials Figure by MIT OCW. ESD.86 Class #2 February 12, 2007 Analyzing a Probability Problem Four Steps to Happiness 1. Define the Random Variable(s) 2. Identify the (joint) sample space 3. Determine the probability law over the ...
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who pass by up to and including the one who is the kth person interviewed. P{Y=y}=P{exactly k-1 interviews occur in (y-1) people passing AND the yth person passing is Negative Binomial Probability Mass Function interviewed} Indicator random variables. Suppose Then E[Xi] = 1*pi + 0*(1- pi) = pi . Example 1: Flip a c...
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= the number of hats returned to the correct owners. Then, Answer independent of the number of players or teams! Example 2. Baseball Hats. Suppose that one player from each of the 30 Major League Baseball teams attends a party at MIT, and each arrives wearing his team’s baseball cap. Each tosses his hat into a closet...
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(1738; English trans. 1954) ‘Google’ this & find many interesting articles, such as http://plato.stanford.edu/entries/paradox-stpetersburg/ Does the St. Petersburg Paradox Occur in Nature? A possible term project! Size and frequency of occurrence of Earthquakes. Small earthquakes occur every day all around the world, ...
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grenade Construction site blast WWII conventional bombs late WWII conventional bombs WWII blockbuster bomb Massive Ordnance Air Blast bomb Chernobyl nu clear disaster, 1986 Small atomic bomb Average tornado (total energy) Nagasaki atomic bomb Little Skull Mtn., NV Quake, 1992 Double Spring Fl at, NV Quake, 1994 Northri...
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Let T12 = event successful Transmission from node 1 to node 2. Assume all links function independently. Then, P{T12}=P{E or (AB) or (AC) or (DB) or (DC)} P{T12}=P{E + E’[(AB) + (AC) + (DB) + (DC)]} P{T12}=P{E + E’[A(B+C) + D(B +C)]} P{T12}=P{E + E’[(A+D)(B+C)]}= P{E + E’[(A+A’D)(B+B’C)]} P{T12}= pE + (1-pE){[pA+(1- ...
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%jacobian2by2.m %Code 8.1 of Random Eigenvalues by Alan Edelman %Experiment: Compute the Jacobian of a 2x2 matrix function %Comment: Symbolic tools are not perfect. The author % exercised care in choosing the variables. syms p q r s a b c d t e1 e2 X=[p q ; r s]; A=[a b;c d]; %% Compute Jacobians Y=X^2; J=jacob...
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Intro Administrivia. • Signup sheet. • prerequisites: 6.046, 6.041/2, ability to do proofs • homework weekly (first next week) • collaboration • independent homeworks • grading requirement • term project • books. • question: scribing? Randomized algorithms: make random choices during run. Main benefits: • speed...
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versarial” may mean “well structured” i.e. natural • fingerprinting/verification – generate short random fingerprints for things – faster than comparing things – almost every fingerprint works – so a random one works 2 • random sampling. graph algs, computational geometry, median – fast way to find “typical” member...
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i + 1) (may have outer elements) • analysis: n n pij ≤ � � i=1 j>i 2/(j − i + 1) � � i=1 j>i n n−i+1 2/k 1/k = � � i=1 k=1 n n ≤ 2 � � i=1 k=1 ≤ 2nHn 4 (Define Hn, claim O(log n).) = O(n log n). • analysis holds for every input, doesn’t assume random input • we proved expected. can show...
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) ≤ 2Hn � u • result: exists size O(n log n) auto • gives randomized construction • equally important, gives probabilistic existence proof of a small BSP • so might hope to find deterministically. MinCut • the problem • contraction • conditionally independent events • give/analyze • repetition for better succ...
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Why to Study Finite Element Analysis! That is, “Why to take 2.092/3” Klaus-Jürgen Bathe Why You Need to Study Finite Element Analysis! Klaus-Jürgen Bathe Analysis is the key to effective design effective design We perform analysis for: • deformations and internal forces/stresses • temperatures and heat tr...
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of finite element computer programs (such as NASTRAN, ANSYS, ADINA, SIMULIA, etc…) These analysis programs are interfaced with computer-aided , ) p g design (CAD) programs Catia, desi n CAD ro rams Catia SolidWorks, Pro/Engineer, NX, etc. g ( The process of modeling for analysis The process of modeling fo...
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. model Use of 3x3 16el. model Use of 2x 2 16x64 element model use of 3x3 Gauss Gauss integration Gauss integration integration 1 1 2 3 4 5 6 112.4 112.4 634.5 906.9 154 8 2654 2691 110.5 110.5 617.8 905.5 958.4 * 1528 2602 110.6 110.6 606.4 905.2 1441 2345 2664 *Spurious mode (phantom o...
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7) • Probably due to the Sleipner accident, • Probabl due to the Slei ner accident , p y increased analysis attention was given to critical components – designers and analysts worked closely together Accuracy - part of reality Coarse Mesh Converged Mesh Reference Mesh Correct surface stress prediction at cr...
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pled with structural interactions – an increasingly important analysis area • Full Navier-Stokes equations for incompressible or fully compressible flows • Arbitrary Lagrangian-Eulerian formulation for the fluid Shock absorber Shock absorber Assembly parts Shock absorber Structural model Shock absorber ...
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Particle trace plot Analysis of an artificial lung Particle trace Radio-frequency tissue ablation Electrode Lesion Courtesy of Medtronic, Inc. Used with permission. Radio-frequency tissue ablation Catheter Electrode Symmetry face Blood Tissue Radio-frequency tissue ablation Temperature variation during...
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ELECTRIC FORCES ON CHARGES Lorentz Force Law: + × μ )o H Newtons a = f/m = qE/m ≈ eV/mL [m s-2] ( E v = q f Kinematics*: t v = a(t)dt ∫ 0 = v o + ˆ at z cathode -V - E⊥ heated filament + z z = z + z•v t + at /2 o ˆ 2 o ⇒ f = qE ma = anode, phosphors deflection plates cathode ray tube ...
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Attractive pressure L5-2 ENERGY METHOD FOR FINDING FORCES Force, work, and energy: dw = f ds ⇒ f = dw ds [N] C = εοA/s 1 w = CV2 = 2 2 2 1 Q s 1 Q = 2 C 2 A ε o [J] = f = Q ≠ f(s) if C is open circuit 2 1 Q dw 2 Aε ds o 2 ( EA) 1( ε 2 o E )A = o 2 A 2 ε o = -PeA [N] 1 E ε ...
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= = 2 ε 2 A = − P A e *C = εA/s L5-4 LATERAL FORCES – ENERGY METHOD Energy derivative: = − f (externally applied) dw dD 2 w = 2 s Q Q 2C 2 WD ε o = W E +Q C A’ = Ws Fringing field s A’ f D -Q = f 2 Q s = 2 2 WD ε o 2 ( EWD) ε o 2 WD ε o 2 s ( = 1 2 2 E ε o )Ws = ...
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2 R ε θ o = 2 E ε o 2 A' [Nm] ≅ pressure × gap-area A' × θ R + stator rotor - A’ = 2Rs R 2 - + T Motor power: Peak power: Average power: Pavg = P/2 (duty cycle = ½) P = Tω [W] n = 4 θ Segmentation advantage: T [Nm] = -dw/dθ ∝ A’ ∝ nRs (n = # segments) v + - rotor stator L5-6 ELEC...
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MIT OpenCourseWare http://ocw.mit.edu 18.917 Topics in Algebraic Topology: The Sullivan Conjecture Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. The Adem Relations (Continued) (Lecture 5) We continue to work with complexes over the finite field F2 with ...
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let’s consider the case where V = F2 is a complex concentrated in degree 0. In this case, we can identify VhΣ2 with the chain complex C (BΣ2), and we can identify V hΣ2 with the cochain complex C ∗(BΣ2). The norm map induces a map ∗ Hn(BΣ2) → H−n(BΣ2). This is just the usual norm map in the theory of group cohomolo...
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, t−1]/F2[t]. ∗ Using this isomorphism, H (BΣ2) has a basis consisting of {tn}n<0. In previous lectures, we used a basis {xi}i≥0 for H (BΣ2) which was dual to the basis {ti}i≥0 for H∗(BΣ2). By comparing degrees, we see that these bases are related by the following transformation ∗ ∗ xi �→ t−i−1 . It follows that the ...
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induced maps → V . We obtain f : D2(F2)[−2n] � D2(F2[−n]) → D2(V ) f � : DT (F2)[−2n] � D2(F2[−n]) → DT (V ). For every integer k, we let Sk(v) ∈ Hn+k(DT (V )) denote the image of tk−n ∈ Hk−n(DT (F2)) under the map f �. If k ≥ n, then tk−n ∈ Hk−n(D2(F2)) ⊆ Hk−n(DT (F2)). In this case, we will denote the image o...
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on V : (∗) The cohomology groups Hn(V ) are finite dimensional for every n ∈ Z, and vanish for n sufficiently small. Assuming condition (∗), we have equivalences V hΣ2 � V ⊗ (F2)hΣ2 V T Σ2 � V ⊗ (F2)T Σ2 VhΣ2 � V ⊗ (F2)hΣ2 . Passing to cohomology, we obtain isomorphisms H∗(V hΣ2 ) � H∗(V )[t] H∗(V T Σ2 ) � H∗(V )[...
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fies tmSk(v) = Sm+k(v). → ⇒ 2 The coefficient of tk−l in φ(Sk(v)) is given by Res(tl−k−1φ(Sk(v))) = Res(φ(Sl−1(v))). 3 � � � � � We have a commutative diagram H∗(V ) Sl−1 � � H∗(DT (V )) � � id H∗( V ) Sql � � H∗(D 2(V )) � H∗(V )[t, t−1 H∗(V )[t, t−1] ������������� Res ]/ H∗(V )[t] Res � H∗(V ). We now ...
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the following: Corollary 4. The inclusion j : Σ2 × Σ2 → G induces a restriction map on cohomology H∗(BG) Σ2) � F2[t, u]. For k ≥ n, this map carries Sk(un) ∈ Hm+k(BG) to → H∗(Σ2 × � (n − l, l)u n+ltk−l . p We observe that H (BG) � H−∗(D2(C (BΣ2))) has a basis consisting of products {xixj }0≤i<j and xi}0≤i≤n. We o...
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complete the calculation of the last lecture. Recall that we need to show that for l p, q > 0, the homology classes � (p − 2l, l) Sq xp−l ∈ Hp+q(BG) −q−l l 4 � � � � � � � � � � � (q − 2l�, l�) Sq −p−l� xq−l� ∈ Hp+q(BG) l� have the same image in H (BΣ4). Invoking Corollary 5, we see that it suffices to show ...
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6.890 Algorithmic Lower Bounds and Hardness Proofs Fall 2014 Erik Demaine Class 2 Scribe Notes 1 Useful Problems for Hardness Reductions This lecture mostly focuses on using 3-Partition to solve 2+ problems by reducing to number problems. The basic idea is to think of your numbers as integers – fixed-point or rati...
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S ⊂ A such that ΣS = t. It is easy to think of an instance of this problem as a partition, although it’s a generalization. Reducing from Subset Sum that we can reduce from 2-Partition. 2-Partition to Subset Sum is a strict generalization – not given t – but we are essentially choosing a subset A1 whose sum is ΣA/2. ...
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A), but all the ai’s become roughly equal to each other and arbitrarily close to t/3. Let’s talk like n about a couple of related problems to 3-Partition, and why we’re talking about it. 100 1.2.1 Numerical 3-Dimensional Matching This problem has a funny name – we will get to that second, don’t worry for the moment....
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× max(A ∪ B ∪ C). The new tf becomes t + 13∞. This forces us to pick one from A, one from B, and one from C. We must show that the infinities can be treated algebraically despite being so large. Note that we require that the sets be of size exactly 3 as part of the 3-Partition specification for this reduction. 1.2.2 ...
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to Num 3-Dim Matching). Because tripartite hypergraphs are 3-uniform, X3C is a strict generalization of 3DM. 1.3 Strong vs. Weak Back to the issue of 2-Partition vs. 3-Partition, we explore an important distinction – weak vs. strong NP-hardness. There are two types of NP-hardness for number problems when we have in...
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But today, we’re going to use unary a lot to talk about Strongly NP-hard problems. So, note: Strong NP-hardness is “My problem is so hard that even if I encode my numbers in unary, it’s still NP-hard.” If you can prove Strong, you should. It’s better. Of course Weak NP-hardness is still okay. 1.4 Pseudopolynomial, ...
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general, strongly NP-hard is a better result, because it is more restrictive. • 2-Partition is weakly NP-hard and Pseudopolynomial (therefore not strongly NP-hard) • 3-Partition is strongly NP-hard (therefore not even Pseudopolynomial) 3 2 Reductions! Let’s do some reductions, shall we? It’s NP-hardcore time! Jus...
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Let’s look at a special case: packings should be exact – no gaps. Rotation must be integer ×90◦, so constant number of them. Translations are integral also given exact packing , so there are succinct encodings implying that this problem is ∈NP. We now prove that both of these problems are strongly NP-hard. 2.2.1 Re...
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n unit squares (tiles), each one with 4 colors (a, b, c, d), with one color per edge; and also a target rectangle, and we want to pack the squares into the rectangle with colors matching. Eternity II, is also an edge matching puzzle, currently still unsolved, which for a while had US $2,000,000 in prize money; this ...
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n/3 t $ $ $ % % % % % 5 Are there any numbers in this problem, as inputs? The colors need to be represented as numbers (although they are only compared, never added). Total number of different colors ≤ 4n, anyways. So it wouldn’t make sense to say this is Strongly NP-hard because the problem can really onl...
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colors are unique pairs to force their assembly. Through this reduction, we see that the hardness is the same for Signed and Unsigned Edge-Matching puzzles. 2.5 Jigsaw Puzzles For Jigsaw Puzzles, each edge is either straight, pocket, or tab. Pockets and tabs can be slightly different in shape. We can allow for some ...
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use log 4n bits along the edge by indent or outdenting accordingly (See Slide 13). They will fit together and make a nice clean seam if the colors match. The construction is blown up by a logarithmic factor log 4n, but the number of pieces stays the same. Now, we reduce Polyomino Packing to Edge-Matching ... by makin...
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of into a rectangle, we create a rectangle gadget (See Slide 14). Make it so that the extra space is a rectangle and then scale up by 3B + t. If you want to completely pack the target square, you can compute how much extra slop there is and add in a bunch of 1x1 tiles so that you can fill in the extra space, making i...
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Chapter 8 Meson Mass Matrix Close up on meson mass matrix states as: Vm = υ + 1 n f √6 )2 mu + + 1 π◦ 8 √2 1 π◦ + 1 a F √3 ( { 1 a F √3 − 8 √2 1 a 1 2n F √3 − f √6 υ 2f 2 a π◦ η ( ( 1 n f √6 )2 ms} � � f 2 2 F 2 3(mu + md + ms) − √6 2f mu+md− F 3√2 2f mu F 2ms md ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎝ a π◦ η ⎞ ⎟ ⎠...
https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2c3cf052ae853f5b31fa7246895672d6_chap8.pdf
+md− F 3√2 1 3(mu − √ 1 + md + 4ms) 3 (mu md) ⎞ ⎟ ⎟ ⎠ 1 (axions). There is a small eigenvalue associated with an f 2 F 2 ∼ eigenvalue ∼ ⎛ ⎜ ⎝ 1 f aF f bF ⎟ ⎞ ⎠ with 60 = = ⎛ ⎜ ⎜ ⎝ � So ∈ M υ 2 ) ( f 1. For f F 61 + (mu + md)a + 1 √3 (mu − md)b = 0(8.4) md 2 mu − √6 1 √3 2ms) + √2 3 (mu...
https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2c3cf052ae853f5b31fa7246895672d6_chap8.pdf
we take mu = md = 0 ∈ M υ ( f 2 ) = ⎛ ⎜ ⎜ ⎝ f 2 F 2 2 3 ms 0 0 0 4 3√2 ms 0 f F − f 4 3√2 ms F − 0 4 3ms ⎞ ⎟ ⎟ ⎠ (8.6) (8.7) (8.8) (8.9) (8.10) has two vanishing eigenvalues. So η gets infected and the GM-O relation is badly violated. The general case is a little messy but with mu = md � ms w...
https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2c3cf052ae853f5b31fa7246895672d6_chap8.pdf
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341: Discrete-Time Signal Processing OpenCourseWare 2006 Lecture 7 IIR, FIR Filter Structures Reading: Sections 6.1 - 6.5 in Oppenheim, Schafer & Buck (OSB). Signal Flow Graphs A linear time-invariant discrete-time ...
https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/2c6fe95a4d44834ccf0c0d9017118703_lec07.pdf
signal glow graph corresponding to the first order system in OSB Figure 6.10. By convention, the delay element has been represented by a branch gain of z−1 . The signal flow graph representation of a LTI system is not unique. In fact, for any given rational system function, equivalent sets of difference equations and ...
https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/2c6fe95a4d44834ccf0c0d9017118703_lec07.pdf
to physical memories in actual implementation, direct form II structures require less state memory than the direct form I implementation. However, the total memory requirement for both forms are similar, because direct form II structures need more cache memory during computations. Transposed Forms Using signal flow...
https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/2c6fe95a4d44834ccf0c0d9017118703_lec07.pdf